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Question: Find the derivative of the following: At \[\left( 1,-1 \right)\]find \[{y}'\]for \[{{x}^{4}}+4{{x}...

Find the derivative of the following:
At (1,−1)\left( 1,-1 \right)find y′{y}'for x4+4x2y3+y2=2y{{x}^{4}}+4{{x}^{2}}{{y}^{3}}+{{y}^{2}}=2y

Explanation

Solution

Hint:If u and v are two differentiable functions of x then ddx(uv)=u×dvdx+v×dudx\dfrac{d}{dx}\left( uv \right)=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}and this formula is called product rule. In this problem u and v are two differentiable functions of x so apply the product rule. If they ask to find the derivative at a particular point substitute the values of x and y in the obtained derivative value.

Complete step-by-step answer:
Given that x4+4x2y3+y2=2y{{x}^{4}}+4{{x}^{2}}{{y}^{3}}+{{y}^{2}}=2y
First apply derivative on both sides with respect to x then we will get
4x3+4(y3×2x+x23y2dydx)+(2y)dydx=2dydx4{{x}^{3}}+4\left( {{y}^{3}}\times 2x+{{x}^{2}}3{{y}^{2}}\dfrac{dy}{dx} \right)+\left( 2y \right)\dfrac{dy}{dx}=2\dfrac{dy}{dx}. . . . . . . . . . . . . . . . . . (1)
4x3+8xy3=dydx(−12x2y2−2y+2)4{{x}^{3}}+8x{{y}^{3}}=\dfrac{dy}{dx}\left( -12{{x}^{2}}{{y}^{2}}-2y+2 \right). . . . . . . . . . . . . . . . . . . . (2)
dydx=4x3+8xy3−12x2y2−2y+2\dfrac{dy}{dx}=\dfrac{4{{x}^{3}}+8x{{y}^{3}}}{-12{{x}^{2}}{{y}^{2}}-2y+2}. . . . . . . . . . . . . . . . . . (3)
The derivative of the following y′{y}'is 4x3+8xy3−12x2y2−2y+2\dfrac{4{{x}^{3}}+8x{{y}^{3}}}{-12{{x}^{2}}{{y}^{2}}-2y+2}
Given they asked us to find the derivative at the point (1,−1)(1,-1)
dydx=4(1)3+8(1)(−1)3−12(1)2(−1)2−2(−1)+2\dfrac{dy}{dx}=\dfrac{4{{\left( 1 \right)}^{3}}+8\left( 1 \right){{\left( -1 \right)}^{3}}}{-12{{\left( 1 \right)}^{2}}{{\left( -1 \right)}^{2}}-2\left( -1 \right)+2}. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
dydx=−4−12+2+2\dfrac{dy}{dx}=\dfrac{-4}{-12+2+2}
dydx=−4−8\dfrac{dy}{dx}=\dfrac{-4}{-8}
dydx=12\dfrac{dy}{dx}=\dfrac{1}{2}
So, the derivative at the point (1,−1)(1,-1)is dydx=12\dfrac{dy}{dx}=\dfrac{1}{2}

Note: The formula for derivative of un=nun−1{{u}^{n}}=n{{u}^{n-1}}where n is any integer and u is any variable like x or y. the x component with respect x is 1 and derivative of y component with respect to x is dydx\dfrac{dy}{dx}as shown this is used in the above problem. Carefully do the basic mathematical operations like addition, subtraction then we will get the required answer.